Annual Appeal: Please make a donation to keep the OEIS running. In 2018 we replaced the server with a faster one, added 20000 new sequences, and reached 7000 citations (often saying "discovered thanks to the OEIS").

See A065091 for comments, formulas etc. concerning only odd primes. For all information concerning prime powers, see A000961. For contributions concerning "almost primes" see A002808.

A number n is prime if it is greater than 1 and has no positive divisors except 1 and n.

A natural number is prime if and only if it has exactly two (positive) divisors.

A prime has exactly one proper positive divisor, 1.

The paper by Kaoru Motose starts as follows: "Let q be a prime divisor of a Mersenne number 2^p-1 where p is prime. Then p is the order of 2 (mod q). Thus p is a divisor of q - 1 and q > p. This shows that there exist infinitely many prime numbers." - Pieter Moree, Oct 14 2004

1 is not a prime, for if the primes included 1, then the factorization of a natural number n into a product of primes would not be unique, since n = n*1.

Prime(n) and pi(n) are inverse functions: A000720(a(n)) = n and a(n) is the least number m such that a(A000720(m)) = a(n). a(A000720(n)) = n if (and only if) n is prime.

A number n is prime if and only if it is different from zero and different from a unit and each multiple of n decomposes into factors such that n divides at least one of the factors. This applies equally to the integers (where a prime has exactly four divisors (the definition of divisors is relaxed such that they can be negative)) and the positive integers (where a prime has exactly two distinct divisors). - Peter Luschny, Oct 09 2012

Motivated by his conjecture on representations of integers by alternating sums of consecutive primes, for any positive integer n, Zhi-Wei Sun conjectured that the polynomial P_n(x)= sum_{k=0}^n a(k+1)*x^k is irreducible over the field of rational numbers with the Galois group S_n, and moreover P_n(x) is irreducible mod a(m) for some m<=n(n+1)/2. It seems that no known criterion on irreduciblity of polynomials implies this conjecture. - Zhi-Wei Sun, Mar 23 2013

I conjecture that for any positive rational number r there are finitely many primes q_1,...,q_k such that r = Sum_{j=1..k} 1/(q_j-1). For example, 2 = 1/(2-1)+1/(3-1)+1/(5-1)+1/(7-1)+1/(13-1) with 2, 3, 5, 7 and 13 all prime, 1/7 = 1/(13-1)+1/(29-1)+1/(43-1) with 13, 29 and 43 all prime, and 5/7 = 1/(3-1)+1/(7-1)+1/(31-1)+1/(71-1) with 3, 7, 31 and 71 all prime. - Zhi-Wei Sun, Sep 09 2015

I also conjecture that for any positive rational number r there are finitely many primes p_1,...,p_k such that r = sum_{j=1..k} 1/(p_j+1). For example, 1 = 1/(2+1)+1/(3+1)+1/(5+1)+1/(7+1)+1/(11+1)+1/(23+1) with 2, 3, 5, 7, 11 and 23 all prime, and 10/11 = 1/(2+1)+1/(3+1)+1/(5+1)+1/(7+1)+1/(43+1)+1/(131+1)+1/(263+1) with 2, 3, 5, 7, 43, 131 and 263 all prime. - Zhi-Wei Sun, Sep 13 2015

a(n) = 1 + Sum_{m=1..L(n)} (abs(n-Pi(m))-abs(n-Pi(m)-1/2)+1/2), where Pi(m) = A000720(m) and L(n) >= a(n)-1. L(n) can be any function of n which satisfies the inequality. For instance, L(n) can be ceiling((n+1)*log((n+1)*log(n+1))) since it satisfies this inequality. - Timothy Hopper, May 30 2015, Jun 16 2015

(PARI) /* The following functions provide asymptotic approximations, one based on the asymptotic formula cited above (slight overestimate for n > 10^8), the other one based on pi(x) ~ li(x) = Ei(log(x)) (slight underestimate): */